Introduction

Now that we know how to differentiate inverse functions, the next logical step is to find the derivatives of some common inverse functions. Specifically the inverse trigonometric functions. These derivatives can be found using the methods we learned earlier in this unit.

Finding the Derivatives of the Inverse Trig Functions

Problem: Find the derivative of arcsin(x)

To find the inverse of trig functions, we can play the same game we did last time and apply the formula $\frac{1}{f’(f^{-1}(x))}=\frac{dy}{dx}$. $f(x)=\sin (x)$, and so applying the formula will give us $\frac{1}{f’({\sin }^1(x))}=\frac{1}{\cos ({\sin }^{-1}(x))}$. This answer does work, but it isn’t very nice. Are you sure you will always be able to find the value of $\cos ({\sin }^{-1}(x))$ without a calculator? Also, this is just a really awkward function, so we will have to do something else.

We can start with $y={\sin }^{-1}(x)$, since that’s the function we’re trying to find. If we were to take the sine of both sides, we can sort of remove the ${\sin }^{-1}(x)$ and get $\sin (y)=x$. This looks a lot cleaner, but now what?

First let's let ${\sin }^{-1}(x)=y$ and $\sin (y)=x$. Then we can differentiate $\sin (y)=x$ implicitly with respect to $x$ to find the derivative of ${\sin }^{-1}(x)$ or $\arcsin (x)$:

$\sin (y)=x$

$\frac{dy}{dx}[\sin (y)]=\frac{dy}{dx}[x]$

$\cos (y)\cdot \frac{dy}{dx}=1$

$\frac{dy}{dx}=\frac{1}{\cos (y)}$

And we’re done! 

Actually, there’s still more. We cannot start with a function that has $x$ and end with a derivative that has $y$, so we need to continue. Well, it would be nice if we could have $\sin (y)$ so that we can use the equation $\sin (y)=x$ that we found earlier to substitute $x$ into and get back the original variable.

In this case, we can use the Pythagorean Identity being ${\sin }^2(y)+{\cos }^2(y)=1$. In this case, $\cos (y)=\sqrt{1-{\sin }^2(y)}$. Since we know that $\sin (y)=x$, we can substitute in $x$ for $\sin (y)$ and then just solve for $\cos (y)$: 

$x^2+{\cos }^2(y)=1$

${\cos }^2(y)=1-x^2$

$\cos (y)=\pm \sqrt{1-x^2}$

Since we are left with two answers for what $\cos (y)$ is equal to, we need to choose which one to use based on the range of arcsin. You may remember from trigonometry that the range of $\arcsin (x)$ is $y\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, and over that range of the $y$ output values for cosine are either always positive or equal to zero. So we will use the positive answer for $\cos (y)=\sqrt{1-x^2}$ and just plug that into the equation we found earlier to get the derivative in terms of $x$:

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

Find the derivative of arccos(x)

The process for finding the derivative of the inverse cosine or arccosine of $x$ is very similar so I will quickly go over the steps for that. Again let's similarly let ${\cos }^{-1}(x)=y$ and $\cos (y)=x$. Then we can differentiate $\cos (y)=x$ implicitly with respect to $x$ to find the derivative of ${\cos }^{-1}(x)$ or $\arccos (x)$:

$\cos (y)=x$

$\frac{dy}{dx}[\cos (y)]=\frac{dy}{dx}[x]$

$-\sin (y)\cdot \frac{dy}{dx}=1$

$\frac{dy}{dx}=-\frac{1}{\sin (y)}$

We now again have the derivative in terms of $y$ rather than $x$. So we need to solve for it in terms of $x$ again:

$\sin (y{)}^2+x^2=1$

$\sin (y{)}^2=1-x^2$

$\sin (y)=\pm \sqrt{1-x^2}$

We again get two answers for our derivative, one negative and one positive. We can again figure out by using the range of the function we are trying to find the derivative of, that being arccos. The range of arccos is $y\in [0,\pi ]$, and over that range the sine function is always non negative, $\sin (y)\ge 0$. So we will choose the positive answer and plug that in:

$\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}$

Find the derivative of arctan(x)/arccot(x)

There are two more trig functions to differentiate, that being ${\tan }^{-1}(x)$ or $\arctan (x)$ and ${\cot }^{-1}(x)$ or $\mathrm{arccot}(x)$ . We will follow a similar process where we let $\arctan (x)=y$ and $\tan (y)=x$ and then differentiate it implicitly:

$\tan (y)=x$

$\frac{dy}{dx}[\tan (y)]=\frac{dy}{dx}[x]$

${\sec }^2(y)\cdot \frac{dy}{dx}=1$

$\frac{dy}{dx}=\frac{1}{{\sec }^2(y)}$

To find this in terms of $x$ we again use a Pythagorean Trigonometric identity. Specifically ${\tan }^2(y)+1={\sec }^2(y)$ and substitute $\tan (y)=x$ to get $x^2+1={\sec }^2(y)$. Now we just plug that in to get:

$\frac{dy}{dx}=\frac{1}{1+x^2}$

We can repeat this process one more time for  $\mathrm{arccot}(x)$ by letting $\mathrm{arccot}(x)=y$ and $\cot (y)=x$:

$\cot (y)=x$

$\frac{dy}{dx}[\cot (y)]=\frac{dy}{dx}[x]$

$-{\csc }^2(y)\cdot \frac{dy}{dx}=1$

$\frac{dy}{dx}=-\frac{1}{{\csc }^2(y)}$

Now we use another Pythagorean Identity, this time ${\csc }^2(y)=1+{\cot }^2(y)$. Substituting we get ${\csc }^2(y)=1+x^2$ and we plug in again for the simplified derivative of $\mathrm{arccot}(x)$:

$\frac{dy}{dx}=-\frac{1}{1+x^2}$

More on the Derivatives of the Inverse Trig Functions

Now that we know the derivatives of $\arcsin (x)$, $\arccos (x)$, $\arctan (x)$, and $\text{arccot}(x)$ you may be wondering about the derivatives of the reciprocal inverse functions. For the purposes of Calculus AB these derivatives are not needed to be known, they are also much more complex to derive so I won’t be doing that here. The derivatives of three main inverse trigonometric functions we just went over are very important to know though:

$\frac{dy}{dx}(\arcsin (x))=\frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}(\arccos (x))=-\frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}(\arctan (x))=\frac{1}{1+x^2}$

$\frac{dy}{dx}(\text{arccot}(x))=-\frac{1}{1+x^2}$

Applying the Derivatives of the Inverse Trig Functions

Find the tangent line of $f(x)=e^{\arctan (2x)}$ at $x=0$

Let’s first solve for the derivative. Just like with any chain rule, let’s list out the inner and outer functions. The inner function is $\arctan (2x)$ and the outer function is $e^x$. The inner function can be $v(x)$ and the outer function can be $u(x)$. This gives us:

$u(x)=e^x$

$v(x)=\arctan (2x)$

$u’(x)=e^x$

To find $v’(x)$, we will need to use the Chain Rule. $\arctan (x)$ is the outer function, and $2x$ is the inner function. 

$\text{Outer}=\arctan (x)$

$\text{Inner}=2x$

$\text{Outer Differentiated}=\frac{1}{1+x^2}$

$\text{Inner Differentiated}=2$

Putting everything together using the chain rule gives us $\frac{1}{1+4x^2}\cdot 2$, or just $\frac{2}{1+4x^2}$. Finally, we can plug in everything into the chain rule 

$u(x)=e^x$

$v(x)=\arctan (2x)$

$u’(x)=e^x$

$v’(x)=\frac{2}{1+4x^2}$

The derivative of the function is $f’(x)=e^{\arctan (2x)}\cdot \frac{2}{1+4x^2}$. To find the tangent line, let’s set $x=0$ and find the slope.

$f’(0)=e^{\arctan (2(0))}\cdot \frac{2}{1+4(0{)}^2}=1\cdot 2=2$. The function at $x=0$ is $f(0)=e^{\arctan (2(0))}=1$. Plugging this all into the point-slope form gives us $y=2(x-0)+1⟹y=2x+1$.

Notes on the Derivatives of the Inverse Trig Functions

Practice Questions